I have this two examples of the cumulative distribution function and I have to find the missing variable $k$.
Example 1: $$ F_X(x)=\begin{cases}0 & \ x \le1 \\\frac{1}{2}x^{2}+kx & 1<x<2 \\1 & 2 \le x \end{cases} $$ I think here we have to equate $\frac{1}{2}x^{2}+kx$ with $1$, but I do not know with what to replace $x$ so I can get $k$.
Example 2: $$ F_X(x)=\begin{cases}0 &x<0 \\\frac{1}{2}x &0\leq x\leq 1 \\1-k(x-4)^{2} &1\leq x\leq 4\\1 &x>4\end{cases} $$ With example 2 I am not sure which one should I equate with $1$.
I would appreciate help solving these two examples.
$F$ is only right-continuous (not necessarily left-continuous), so you should only equate with the value at the right.
$F(2-)=F(2)$, so $\left.\frac{1}{2}x^{2}+kx\right|_{x=2}=1$
and for the second example:
$F(1-)=F(1)$, so $=\left.\frac{1}{2}x\right|_{x=1}=\left.1-k(x-4)^{2}\right|_{x=1}$
Note that $F(4-)=F(4)=1$, so the right-continuity condition in $x=4$ is of no help in finding $k$.